Exponential Covering Map and Its Deck Transformations

The Circle as a Quotient of the Real Line

As a space, the circle can be described by angles, with angles that differ by a full turn representing the same point. The quotient \(\mathbb{R}/\mathbb{Z}\) makes this description precise: it identifies real numbers whose difference is an integer. The exponential map then turns each equivalence class into a point of the unit circle. We will verify the quotient description, check the covering property directly, and determine all the transformations of the covering space that leave the circle unchanged.

Give \(\mathbb{R}\) its standard topology and let \(S^1=\{z\in\mathbb{C}:|z|=1\}\) have the subspace topology inherited from \(\mathbb{C}\). Define an equivalence relation on \(\mathbb{R}\) by \(s\sim t\) if and only if \(s-t\in\mathbb{Z}\). Write \([t]\) for the equivalence class of \(t\), and let \(q:\mathbb{R}\to\mathbb{R}/\mathbb{Z}\) be the map \(q(t)=[t]\). Give \(\mathbb{R}/\mathbb{Z}\) the quotient topology: \(V\subseteq\mathbb{R}/\mathbb{Z}\) is open exactly when \(q^{-1}(V)\) is open in \(\mathbb{R}\).

Consider the map \(p:\mathbb{R}\to S^1\) given by \(p(t)=e^{2\pi i t}\). The elementary angle parametrization of the unit circle gives \(e^{2\pi i t}=\cos(2\pi t)+i\sin(2\pi t)\), and every point of \(S^1\) has this form. Also, \(e^{2\pi i u}=1\) exactly when \(u\in\mathbb{Z}\): the sine is zero and the cosine is one exactly at integer multiples of \(2\pi\). Therefore \(p(s)=p(t)\) exactly when \(s-t\in\mathbb{Z}\). In particular, \(p\) is surjective and is constant on each equivalence class.

Proof. If \([s]=[t]\), then \(s-t\in\mathbb{Z}\), so \(e^{2\pi i s}=e^{2\pi i t}\). Thus \(E\) is well-defined. The equality criterion for \(p\) also shows that \(E\) is injective, and the angle parametrization shows that it is surjective. Since \(p=E\circ q\) is continuous, the definition of the quotient topology implies that \(E\) is continuous: for every open \(W\subseteq S^1\), \(q^{-1}(E^{-1}(W))=p^{-1}(W)\) is open, so \(E^{-1}(W)\) is open in \(\mathbb{R}/\mathbb{Z}\).

To see that \(E\) is open, first note that \(p\) is open. If \(O\subseteq\mathbb{R}\) is open and \(p(t)\in p(O)\), choose \(t\in O\) with that image, and choose \(0<\delta<1/2\) such that \((t-\delta,t+\delta)\subseteq O\). Its image is an open arc of \(S^1\), so \(p(O)\) contains an open neighborhood of each of its points and is open. Now let \(V\) be open in \(\mathbb{R}/\mathbb{Z}\). Then \(q^{-1}(V)\) is open and \(p(q^{-1}(V))=E(V)\), so \(E(V)\) is open. A continuous, open bijection has continuous inverse; hence \(E\) is a homeomorphism. \(\square\)

This identifies the circle with \(\mathbb{R}/\mathbb{Z}\), not merely as a set but with the topology specified by the quotient map. Under this identification, the covering map \(p\) is the quotient map followed by \(E\).

Checking the Covering Property

We verify the covering condition by describing neighborhoods and their full preimages. For \(t_0\in\mathbb{R}\) and \(0<\delta<1/2\), let \(U=p((t_0-\delta,t_0+\delta))\). This is an open arc in \(S^1\). For example, it is the arc obtained by varying the angle \(2\pi t\) strictly between \(2\pi(t_0-\delta)\) and \(2\pi(t_0+\delta)\); its endpoints are excluded.

Proof. Fix \(z_0\in S^1\), and choose \(t_0\in\mathbb{R}\) with \(p(t_0)=z_0\). Choose \(0<\delta<1/2\) and the open arc \(U=p((t_0-\delta,t_0+\delta))\). The equality criterion \(p(s)=p(t)\) if and only if \(s-t\in\mathbb{Z}\) gives \(p^{-1}(U)=\bigcup_{n\in\mathbb{Z}}(t_0+n-\delta,t_0+n+\delta)\). Indeed, a point \(s\) lies in this preimage exactly when \(p(s)=p(r)\) for some \(r\in(t_0-\delta,t_0+\delta)\), which holds exactly when \(s=r+n\) for some integer \(n\).

These intervals are pairwise disjoint: their centers differ by a nonzero integer, whose absolute value is at least \(1\), while each interval has radius \(\delta<1/2\), so its diameter \(2\delta\) is less than \(1\). On each interval \(J_n=(t_0+n-\delta,t_0+n+\delta)\), the restriction of \(p\) is onto \(U\), since \(p(t+n)=p(t)\). It is one-to-one there, since two distinct points in \(J_n\) differ by less than \(2\delta<1\) and therefore cannot differ by a nonzero integer. The restriction is continuous and open: every relatively open subset of \(J_n\) is open in \(\mathbb{R}\), because \(J_n\) is open, and \(p\) maps open sets to open sets. It is consequently a homeomorphism onto \(U\). Thus \(U\) is evenly covered as required. This works for every \(z_0\in S^1\), proving that \(p\) is a covering map. \(\square\)

Fibers and Deck Transformations

Proof. For any \(s\in\mathbb{R}\), \(p(s)=z=p(t_0)\) if and only if \(p(s)=p(t_0)\). By the equality criterion for the exponential map, this holds if and only if \(s-t_0\in\mathbb{Z}\), or equivalently \(s=t_0+n\) for some \(n\in\mathbb{Z}\). This proves both containments in the stated fiber formula. \(\square\)

The fiber formula describes the points that a deck transformation is allowed to move a point to. The next result shows that its choice cannot vary from one point of \(\mathbb{R}\) to another.

Proof. Let \(h\) be a deck transformation. The equation \(p\circ h=p\) implies \(p(h(t))=p(t)\) for every \(t\). By the equality criterion, \(h(t)-t\in\mathbb{Z}\) for every \(t\). The function \(g(t)=h(t)-t\) is continuous, since \(h\) and the identity function are continuous. If there were \(a,b\in\mathbb{R}\) with \(g(a)\ne g(b)\), then \(g(a)\) and \(g(b)\) would be distinct integers. The intermediate value theorem would force \(g\) to take a noninteger value strictly between them, contradicting that \(g(t)\in\mathbb{Z}\) for every \(t\). Hence \(g\) is a constant integer \(n\), and \(h(t)=t+n\) for every \(t\).

Conversely, for each \(n\in\mathbb{Z}\), \(T_n(t)=t+n\) is a homeomorphism, with inverse \(T_{-n}\). Also \(p(T_n(t))=e^{2\pi i(t+n)}=e^{2\pi it}e^{2\pi in}=p(t)\), because \(e^{2\pi in}=1\). Thus \(T_n\) is a deck transformation. The two directions establish the classification. \(\square\)

Composition satisfies \(T_m\circ T_n=T_{m+n}\), since \((T_m\circ T_n)(t)=T_m(t+n)=t+n+m\). Also \(T_0\) is the identity and \(T_n^{-1}=T_{-n}\). Thus the deck transformations form a group isomorphic to \(\mathbb{Z}\), with addition corresponding to composition.

A Common Pitfall

A tempting but incorrect argument is: since \(p(h(t))=p(t)\), choose an integer \(n_t\) for each \(t\) and conclude \(h(t)=t+n_t\); therefore any point-by-point choice of integers defines a deck transformation. The first statement is valid, but the conclusion ignores continuity. For example, set \(h(t)=t\) for \(t<0\), and \(h(t)=t+1\) for \(t\ge0\). Then \(p(h(t))=p(t)\) on both pieces, because adding \(1\) leaves \(p\) unchanged. But \(h\) is discontinuous at \(0\): for \(t<0\) approaching \(0\), \(h(t)\) approaches \(0\), while \(h(0)=1\). It is not a deck transformation. The proof using the intermediate value theorem is what rules out this point-dependent choice and forces one fixed integer translation.

The quotient description, fiber calculation, and deck-transformation classification express the same periodicity in three forms: real numbers are identified modulo integers, each fiber is an integer translate of one point, and every symmetry of the cover over the circle is an integer translation. The covering property supplies the local structure that makes this periodic map a model example for covering-space arguments.